GR - Lie Derivative of metric - Killing Equation

[binbagsss]

Homework Statement

Question attached.

Homework Equations

3. The Attempt at a Solution [/B]

I'm not really sure how to work with what is given in the question without introducing my knowledge on lie derivatives.

We have: $(L_ug)_{uv} = U^{\alpha}\nabla_{\alpha}g_{uv}+g_{u\alpha}\nabla_vU^{\alpha}+g_{\alpha v}\nabla_u U^{\alpha}$

Where $L_u$ denotes the lie derivative in the direction of $u$

Where the first term vanishes on the assumption of a levi-civita connection by the fundamental theorem of riemmanian geometry. $U^{\alpha}\nabla_{\alpha}g_{uv}= U^{\alpha}(\partial_{\alpha}g_{uv} + \Gamma^{c}_{\alpha v} g_{uc} + \Gamma^{c}_{\alpha u} g_{vc})$

And so the fact that $w^u \partial_u g_{ab} =0 \implies \partial_t g_{ab} =0$ vanishes here, implies that the connection terms vanish (here I need an argument that they vanish individually, and that the sum can not vanish) I can then conclude that this implies we are working in Minkowski space-time.

And I can then use my the following knowledge on Lie derivatives to obtain an answer:

$w^u \partial_u g_{ab} =0 \implies \partial_t g_{ab} =0$ ; i.e. the metric has no $t$ dependence so I know that this means that the Lie derivative (1) vanishes:

And then I have

$(L_ug)_{uv} =0= g_{u\alpha}\nabla_vU^{\alpha}+g_{\alpha v}\nabla_u U^{\alpha}=\nabla_vU_u+\nabla_u U_v$

$\implies k=-1$

However the question makes no reference to the requirement of needing Lie derivative, so I'm not too sure about what I've done here,

Can I start from first principles more? Any hint appreciated.

Thanks in advance.

 

[Dazed&Confused]

Here's what I think is one way. Consider $$\nabla_\mu W_\nu = \partial_\mu W_\nu -\Gamma^s_{\nu \mu} W_s = \partial_\mu W_\nu -\tfrac12 W_s g^{s \sigma} ( \partial_\mu g_{\sigma \nu } + \partial_\nu g_{\sigma \mu} - \partial_\sigma g_{\nu \mu}) = \partial_\mu W_\nu -\tfrac12 W^\sigma ( \partial_\mu g_{\sigma \nu } + \partial_\nu g_{\sigma \mu} - \partial_\sigma g_{\nu \mu}) = \partial_\mu W_\nu -\tfrac12 W^\sigma ( \partial_\mu g_{\sigma \nu } + \partial_\nu g_{\sigma \mu})$$
where the last term vanishes by assumption. Now since $W^\mu = (1,0,0,0)$, $W_\mu = g_{\nu \mu} W^\nu = g_{0 \mu}$. Therefore the equation can be rewritten as $$\partial_\mu W_\nu -\tfrac12 W^\sigma ( \partial_\mu g_{\sigma \nu } + \partial_\nu g_{\sigma \mu}) = \partial_\mu g_{0 \nu} - \tfrac12 ( \partial_\mu g_{0 \nu} + \partial_\nu g_{0 \mu}) = \tfrac12 \partial_\mu g_{0 \nu} - \tfrac12 \partial_\nu g_{0 \mu}.$$ It is easy to check that $\nabla_\nu W_\mu = -(\tfrac12 \partial_\mu g_{0 \nu} - \tfrac12 \partial_\nu g_{0 \mu})$, so $k=-1$ as expected.